1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find volume between y-10 =x and y^2 - 6y =x, rotated around x=1. Did I do this right?

  1. Feb 3, 2013 #1
    1. Find the volume between y-10=x and y2 -6y =x, rotated around x=1.


    2. R= y2 -6y -1

    r= y-10 -1


    ∫2 to 5 of [([itex]\pi[/itex](y2 -6y -1)2) - [itex]\pi[/itex](y - 11)2]

    ∫[itex]\pi[/itex][y4 - 12y3 + 4y2 + 12y + 1] - [itex]\pi[/itex][y2 - 22y + 121]dy

    = [itex]\pi[/itex][y5/5 - 12y4/4 + 4y3/3 +122/2 +y] - [itex]\pi[/itex][y3/3 - 22y2/2 + 121y] |2 to 5

    ...LONG SUBSTITUTION...

    = -928.33[itex]\pi[/itex] - 371.67[itex]\pi[/itex] + 4.93[itex]\pi[/itex] + 200.67[itex]\pi[/itex]

    = -1094.4[itex]\pi[/itex]

    ...Is this the right answer...? I don't really think it's right because it's negative!
    But what am I doing wrong here? :(

    Thanks SO much for your help! :D
     
  2. jcsd
  3. Feb 3, 2013 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Re: Find volume between y-10 =x and y^2 - 6y =x, rotated around x=1. Did I do this ri

    It's easy to get something negative if you are subtracting something larger from something smaller when you meant to do the reverse. Which of those curves has the largest radius around the axis x=1? And I don't think just flipping the sign right will fix it either. Some other error around as well. Ah, you didn't square y^2-6y-1 correctly.
     
  4. Feb 4, 2013 #3
    Re: Find volume between y-10 =x and y^2 - 6y =x, rotated around x=1. Did I do this ri

    You're right! I didn't square the (y2 - 6y -1) correctly! X(

    The correct sqaring should equal: (y4 - 12y3 + 34y2 + 12y + 1)

    So the integral seems like is should really be:

    ∫2 to 5 [[itex]\pi[/itex](y4 - 12y3 + 34y2 + 12y + 1) - [itex]\pi[/itex](y2 - 22y + 121)]dy

    AWFUL AND LONG SUBSTITUTION!

    = 321.67[itex]\pi[/itex] - 371.67[itex]\pi[/itex] - 75.07[itex]\pi[/itex] + 199.33[itex]\pi[/itex]

    = 74.26[itex]\pi[/itex]

    BUT when I put this whole integral into Wolfram Alpha (including the squares), it said it equals 0!

    Then when I put in the integral again (this time with the squares factored out), it said the answer should be 15.9016!
    What?! Would you please tell me what's wrong here?

    Thank you SO much! :D
     
  5. Feb 4, 2013 #4
    Re: Find volume between y-10 =x and y^2 - 6y =x, rotated around x=1. Did I do this ri

    I went through this thing again and again, and I only thing I found was a slight error making the answer 75.6[itex]\pi[/itex]!

    Do you guys have any idea about this??? :(
     
  6. Feb 4, 2013 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Re: Find volume between y-10 =x and y^2 - 6y =x, rotated around x=1. Did I do this ri

    I think 75.6*pi sounds just fine. You KNOW 0 isn't correct, right? What exactly did you input to WA?
     
  7. Feb 5, 2013 #6
    Re: Find volume between y-10 =x and y^2 - 6y =x, rotated around x=1. Did I do this ri

    integral from 2 to 5 of [pi(y^4 - 12y^3 + 34y^2 + 12y + 1) - pi(y^2 - 22y + 121)]


    The above is exactly what I typed into WA.
     
  8. Feb 5, 2013 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Re: Find volume between y-10 =x and y^2 - 6y =x, rotated around x=1. Did I do this ri

    That's an interesting misunderstanding. WA takes pi(x) to be a function giving the number of primes less than or equal to x. http://en.wikipedia.org/wiki/Prime-counting_function Try reading the fine print in the input box. If don't want that write pi*(x) instead.
     
    Last edited: Feb 5, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook